So in order to prove this I am asked to first start off by proving fa ≡ fb(mod m) and fa+1 ≡fb+1 (mod m). The question doesn't specify much. I have an idea on how to prove that the residues are periodic, but I'm not sure how to prove that if we take fa (mod m) we will get a Fibonacci number fb.
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See https://math.stackexchange.com/questions/631171/proof-that-fibonacci-sequence-modulo-m-is-periodic – Robert Z Mar 04 '22 at 16:33
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There are only finite many possibilities for the pair $(a,b)$ modulo $m$. Therefore the sequence must eventually get periodic. – Peter Mar 04 '22 at 16:33
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I already looked at that my question is that how to prove that taking a Fibonacci number mod m we will get another Fibonacci number. Would it be suffice to show it work for m = 3, 4, 5 as the remainders will be 0, 1, 1, 2, 0 for mod 3, 0, 1, 1, 2, 3 for mod m.... and these are all Fibonacci numbers? – R314 Mar 04 '22 at 16:36